This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Texas State University—San Marcos
Ingram School of Engineering EE3340: Fields and Waves
$ 0 L. u T i o u 5 Practice Problem 8: Gauss’s Law, Divergence, Potential
NAME: DATE: For this problem, you will use Gauss’s law and the concept of divergence to solve
the same problem two different ways. Then you will ﬁnd the potential everywhere. Assume there is a slab of charged material whose relative dielectric constant e, =
2.8, and whose uniform volume charge density pv = +1 mC M3. The slab extends to
inﬁnity in the y and z—directions, has a thickness T: 1 cm, and is centered on the x = 0
plane: \>\\ i X yl‘l \\\o :44 1. Using Gauss’s Law and symmetry arguments, ﬁnd the electric ﬁeld E  ii x E x (x)
everywhere. (Since the slab is uniform and inﬁnite in y and 2, there is no Ey or E, ﬁeld.)
For your volume, imagine a cylindrical “pillbox” with an axis in the x—direction and put
its ends at +x and x. Again by symmetry, the ﬂux coming out each end has to be equal
in magnitude but opposite in direction (sign). You will need three expressions for Ex:
one for x < T/2, one for T/2 < x < +T/2, and one for x > T/2. When you are done, sketch
your Ex(x) on the graph provided. Q5 3 L n“ ( (0’1“) 1(1.%)(4.i5¥..°~u”,,} : 2.0 l lav/M aux
. By 6x
symmetry, there is no E)c at the center of the slab (x = 0). Find Dx and then E, by integrating the divergence equation from x = 0 to +x and x values. Your result should be
thesamethatyougotinPartl. x>°1 6 E 35* _. 54E _\___3_C_5_;3__
. ' ° T ‘ I“ "J
x ‘0. sureSg‘“ +OK: “1 2. Now, ﬁnd Ex(x) using the onedimensional divergence equation Pv  € 21 2 *03‘ ‘(0‘ ”/01
1 "l ,
$0 Comte” n :4 1 / + Er: KHO‘B'W/n .9" 0<x<I
E» ‘ “MW“ k"; F" x> x A)». El: (our  vow:
~ . E :. 1onv/nr lfOAN‘ﬁK 7") 0.? I__ _ +zolw/H " 3. Finally, using the onedimensional form of the equation E  VV, ﬁnd the potential
(voltage) V(x) everywhere. Take x: O as the reference point for zero potential (ground)
so that V(O) = 0. Do this by integrating from 0 toward positive and negative values of x.
Again, you will need expressions for three regions (one for x < —T/2, one for T/2 < x <
+T/2, and one for x > T/2). Sketch your result on the graph provided. 9 v V (XI 0
04m 1 o Mk 1 ”1111411ij x
xi S:L‘f 3 / )x 6 0  POMBOL‘ K' W
Z~W03vaﬁ 3)
; T
"H‘ 2, W13: _\vo.)uv.Q.s1o M)"
k, L :—.SO3V luau» >031. 9.41m tum77;) +u¢[email protected]»v(u)4;
.____ 9 [‘;.va cw ”QC?“ wtik 5b” ~10V~Vl ‘9' Li‘ukq
+ i kv/01S‘K as ‘ ‘ ésXHS‘H""') ...
View
Full Document
 Spring '11
 Stephen

Click to edit the document details